Q: Show that the function $f: R_{\bullet} \rightarrow R_{\bullet}$ defined by $(x)=\frac{1}{x}$ is one-one and onto, where $R_{\bullet}$ is the set of all non -zero real numbers. Is the result true, if the domain $R_{\bullet}$ is replaced by N with codomain being same as $R_{\bullet}$ ?

$f: R_{.} \rightarrow R_{.}$is by $f(x)=\frac{1}{x}$ For one-one: $x, y \in R_{\bullet}$ such that $f(x)=f(y)$ $\Rightarrow \frac{1}{x}=\frac{1}{y}$ $\Rightarrow x=y$ $\therefore f$ is one-one. For onto: For $y \in R$, there exists $x=\frac{1}{y} \in R_{.}[$as $y \notin 0]$ such that $f(x)=\frac{1}{\left(\frac{1}{y}\right)}=y$ $\therefore f$ is onto. Given function $f$ is one-one and onto. Consider function $g: N \rightarrow R_{\bullet}$.defined by $g(x)=\frac{1}{x}$ We have, $g\left(x_{1}\right)=g\left(x_{2}\right) \Rightarrow \frac{1}{x_{1}}=\frac{1}{x_{2}} \Rightarrow x_{1}=x_{2}$ $\therefore g$ is one-one. $g$ is not onto as for $1.2 \in R_{\text {there exist any }} x$ in $N$ such that $g(x)=\frac{1}{1.2}$ Function $g$ is one-one but not onto.

Q: Check the injectivity and surjectivity of the following functions: i. $\quad f: N \rightarrow N$ given by $f(x)=x^{2}$ ii. $\quad f: Z \rightarrow Z$ given by $f(x)=x^{2}$ iii. $\quad f: R \rightarrow R$ given by $f(x)=x^{2}$ iv. $\quad f: N \rightarrow N$ given by $f(x)=x^{3}$ v. $\quad f: Z \rightarrow Z$ given by $f(x)=x^{3}$

i. For $f: N \rightarrow N$ given by $f(x)=x^{2} x, y \in N f(x)=f(y) \Rightarrow x^{2}=y^{2} \Rightarrow x=y$ $\therefore f$ is injective. $2 \in N$. But, there does not exist any $x$ in $N$ such that $f(x)=x^{2}=2$ $\therefore f$ is not surjective Function $f$ is injective but not surjective. ii. $\quad f: Z \rightarrow Z$ given by $f(x)=x^{2}$ $f(-1)=f(1)=1$ but $-1 \neq 1$ $\therefore f$ is not injective. $-2 \in Z$ But, there does not exist any $x \in Z$ such that $f(x)=-2 \Rightarrow x^{2}=-2$ $\therefore f$ is not surjective. Function $f$ is neither injective nor surjective. iii. $\quad f: R \rightarrow R$ given by $f(x)=x^{2}$ $f(-1)=f(1)=1$ but $-1 \neq 1$ $\therefore f$ is not injective. $-2 \in Z$ But, there does not exist any $x \in Z$ such that $f(x)=-2 \Rightarrow x^{2}=-2$ $\therefore f$ is not surjective. Function $f$ is neither injective nor surjective. iv. $\quad f: N \rightarrow N$ given by $f(x)=x^{3}$ $x, y \in N$ $f(x)=f(y) \Rightarrow x^{3}=y^{3} \Rightarrow x=y$ $\therefore f$ is injective. $2 \in N$. But, there does not exist any $x$ in $N$ such that $f(x)=x^{3}=2$ $\therefore f$ is not surjective Function $f$ is injective but not surjective. v. $\quad f: Z \rightarrow Z$ given by $f(x)=x^{3}$ $x, y \in Z$ $f(x)=f(y) \Rightarrow x^{3}=y^{3} \Rightarrow x=y$ $\therefore f$ is injective. $2 \in Z$. But, there does not exist any $x$ in $Z$ such that $f(x)=x^{3}=2$ $\therefore f$ is not surjective. Function $f$ is injective but not surjective.

Q: Prove that the greatest integer function $f: R \rightarrow R$ given by $f(x)=[x]$ is neither one-one nor onto, where $[x]$ denotes the greatest integer less than or equal to $x$.

$f: R \rightarrow R$ given by $f(x)=[x]$ $f(1.2)=[1.2]=1, f(1.9)=[1.9]=1$ $\therefore f(1.2)=f(1.9)$, but $1.2 \neq 1.9$ $\therefore f$ is not one-one. Consider $0.7 \in R$ $f(x)=[x]$ is an integer. There does not exist any element $x \in R$ such that $f(x)=0.7$ $\therefore f$ is not onto. The greatest integer function is neither one-one nor onto.

Q: Show that the modulus function $f: R \rightarrow R$ given by $f(x)=|x|$ is neither one-one nor onto, where $|x|$ is $x$, if $x$ is positive or 0 and $|x|$ is $-x$, if $x$ is negative.

$f: R \rightarrow R$ is $$ f(x)=|x|=\left\{\begin{array}{l} \mathrm{x}, \text { if } \mathrm{x} \geq 0 \\ -x, \text { if } \mathrm{x}<0 \end{array}\right\} $$ $f(-1)=|-1|=1$ and $f(1)=|1|=1$ $\therefore f(-1)=f(1)$ but $-1 \neq 1$ $\therefore f$ is not one-one. Consider $-1 \in R$ $f(x)=|x|$ is non-negative. There exist any element $x$ in domain $R$ such that $f(x)=|x|=-1$ $\therefore f$ is not onto. The modulus function is neither one-one nor onto.

Q: Show that the signum function $f: R \rightarrow R$ given by $\quad f(x)=\left\{\begin{array}{l}1, \text { if } \mathrm{x}>0 \\ 0, \text { if } \mathrm{x}=0 \\ -1, \text { if } \mathrm{x}<0\end{array}\right\}$ is neither one-one nor onto.

$f: R \rightarrow R$ is $\quad f(x)=\left\{\begin{array}{l}1, \text { if } \mathrm{x}>0 \\ 0, \text { if } \mathrm{x}=0 \\ -1, \text { if } \mathrm{x}<0\end{array}\right\}$ $f(1)=f(2)=1$, but $1 \neq 2$ $\therefore f$ is not one-one. $f(x)$ takes only 3 values $(1,0,-1)$ for the element -2 in co-domain R , there does not exist any $x$ in domain R such that $f(x)=-2$. $\therefore f$ is not onto. The signum function is neither one-one nor onto.

Q: Let $A=\{1,2,3\}, B=\{4,5,6,7\}$ and let $f=\{(1,4),(2,5),(3,6)\}$ be a function from $A$ to B. Show that f is one-one.

$A=\{1,2,3\}, B=\{4,5,6,7\}$ $f: A \rightarrow B$ is defined as $f=\{(1,4),(2,5),(3,6)\}$ $\therefore f(1)=4, f(2)=5, f(3)=6$ It is seen that the images of distinct elements of $A$ under $f$ are distinct. $\therefore f$ is one-one.

Q: In each of the following cases, state whether the function is one-one, onto or bijective. Justify your answer. i. $\quad f: R \rightarrow R$ defined by $f(x)=3-4 x$ ii. $\quad f: R \rightarrow R$ defined by $f(x)=1+x^{2}$

i. $\quad f: R \rightarrow R$ defined by $f(x)=3-4 x$ $x_{1}, x_{2} \in R$ such that $f\left(x_{1}\right)=f\left(x_{2}\right)$ $\Rightarrow 3-4 x_{1}=3-4 x_{2}$ $\Rightarrow-4 x=-4 x_{2}$ $\Rightarrow x_{1}=x_{2}$ $\therefore f$ is one-one. For any real number $(y)_{\text {in }} R$, there exists $\frac{3-y}{4}$ in $R$ such that $f\left(\frac{3-y}{4}\right)=3-4\left(\frac{3-y}{4}\right)=y \therefore f$ is onto. Hence, $f$ is bijective. ii. $\quad f: R \rightarrow R$ defined by $f(x)=1+x^{2}$ $x_{1}, x_{2} \in R$ such that $f\left(x_{1}\right)=f\left(x_{2}\right)$ $\Rightarrow 1+x_{1}{ }^{2}=1+x_{2}{ }^{2}$ $\Rightarrow x_{1}{ }^{2}=x_{2}{ }^{2}$ $\Rightarrow x_{1}= \pm x_{2}$ $\therefore f\left(x_{1}\right)=f\left(x_{2}\right)$ does not imply that $x_{1}=x_{2}$ Consider $f(1)=f(-1)=2$ $\therefore f$ is not one-one. Consider an element -2 in co domain $R$. It is seen that $f(x)=1+x^{2}$ is positive for all $x \in R$. $\therefore f$ is not onto. Hence, $f$ is neither one-one nor onto.

Q: Let $A$ and $B$ be sets. Show that $f: A \times B \rightarrow B \times A$ such that $(a, b)=(b, a)$ is a bijective function.

$f: A \times B \rightarrow B \times A$ is defined as $(a, b)=(b, a)$. $\left(a_{1}, b_{1}\right),\left(a_{2}, b_{2}\right) \in A \times B$ such that $f\left(a_{1}, b_{1}\right)=f\left(a_{2}, b_{2}\right)$ $\Rightarrow\left(b_{1}, a_{1}\right)=\left(b_{2}, a_{2}\right)$ $\Rightarrow b_{1}=b_{2}$ and $a_{1}=a_{2}$ $\Rightarrow\left(a_{1}, b_{1}\right)=\left(a_{2}, b_{2}\right)$ $\therefore f$ is one-one. $(b, a) \in B \times A$ there exist $(a, b) \in A \times B$ such that $f(a, b)=(b, a)$ $\therefore f$ is onto. $f$ is bijective.

Q: $$ f(n)=\left\{\begin{array}{l} \frac{n+1}{2}, \text { if } n \text { is odd } \\ \frac{n}{2}, \text { if } n \text { is even } \end{array}\right\} $$ Let $f: N \rightarrow N$ be defined as for all $n \in N$. State whether the function $f$ is bijective. Justify your answer.

$f: N \rightarrow N$ be defined as $f(n)=\left\{\begin{array}{l}\frac{n+1}{2}, \text { if } n \text { is odd } \\ \frac{n}{2}, \text { if } n \text { is even }\end{array}\right\}$ for all $n \in N$. $f(1)=\frac{1+1}{2}=1$ and $f(2)=\frac{2}{2}=1$ $f(1)=f(2)$, where $1 \neq 2$ $\therefore f$ is not one-one. Consider a natural number $n$ in co domain $N$. Case I: $n$ is odd $\therefore n=2 r+1$ for some $r \in N$ there exists $4 r+1 \in N$ such that $f(4 r+1)=\frac{4 r+1+1}{2}=2 r+1$ Case II: $n$ is even $\therefore n=2 r$ for some $r \in N$ there exists $4 r \in N$ such that $f(4 r)=\frac{4 r}{2}=2 r$ $\therefore f$ is onto. $f$ is not a bijective function.

Question 1

Let $A=R-\{3\}, B=R-\{1\}$ and $f: A \rightarrow B$ defined by $f(x)=\left(\frac{x-2}{x-3}\right)$. Is $f$ one-one and onto? Justify your answer.

Accepted Answer

$A=R-\{3\}, B=R-\{1\}$ and $f: A ightarrow B$ defined by $f(x)=\left(\frac{x-2}{x-3}ight)$

$x, y \in A_{	ext {such }}$ that $f(x)=f(y)$

$\Rightarrow \frac{x-2}{x-3}=\frac{y-2}{y-3}$

$\Rightarrow(x-2)(y-3)=(y-2)(x-3)$

$\Rightarrow x y-3 x-2 y+6=x y-3 y-2 x+6$

$\Rightarrow-3 x-2 y=-3 y-2 x$

$\Rightarrow 3 x-2 x=3 y-2 y$

$\Rightarrow x=y$

$	herefore f$ is one-one.

Let $y \in B=R-\{1\}$, then $y 
eq 1$

The function $f$ is onto if there exists $x \in A$ such that $f(x)=y$.

Now,

$f(x)=y$

$\Rightarrow \frac{x-2}{x-3}=y$

$\Rightarrow x-2=x y-3 y$

$\Rightarrow x(1-y)=-3 y+2$

$\Rightarrow x=\frac{2-3 y}{1-y} \in A$

$$

[y 
eq 1]

$$

Thus, for any $y \in B$, there exists $\frac{2-3 y}{1-y} \in A$ such that

$f\left(\frac{2-3 y}{1-y}ight)=\frac{\left(\frac{2-3 y}{1-y}ight)-2}{\left(\frac{2-3 y}{1--y}ight)-3}=\frac{2-3 y-2+2 y}{2-3 y-3+3 y}=\frac{-y}{-1}=y$

$	herefore f$ is onto.

Hence, the function is one-one and onto.

Question 2

Show that the function $f: R_{\bullet} \rightarrow R_{\bullet}$ defined by $(x)=\frac{1}{x}$ is one-one and onto, where $R_{\bullet}$ is the set of all non -zero real numbers. Is the result true, if the domain $R_{\bullet}$ is replaced by N with codomain being same as $R_{\bullet}$ ?

Accepted Answer

$f: R_{.} ightarrow R_{.}$is by $f(x)=\frac{1}{x}$

For one-one:

$x, y \in R_{\bullet}$ such that $f(x)=f(y)$

$\Rightarrow \frac{1}{x}=\frac{1}{y}$

$\Rightarrow x=y$

$	herefore f$ is one-one.

For onto:

For $y \in R$, there exists $x=\frac{1}{y} \in R_{.}[$as $y 
otin 0]$ such that $f(x)=\frac{1}{\left(\frac{1}{y}ight)}=y$

$	herefore f$ is onto.

Given function $f$ is one-one and onto.

Consider function $g: N ightarrow R_{\bullet}$.defined by $g(x)=\frac{1}{x}$

We have, $g\left(x_{1}ight)=g\left(x_{2}ight) \Rightarrow \frac{1}{x_{1}}=\frac{1}{x_{2}} \Rightarrow x_{1}=x_{2}$

$	herefore g$ is one-one.

$g$ is not onto as for $1.2 \in R_{	ext {there exist any }} x$ in $N$ such that $g(x)=\frac{1}{1.2}$

Function $g$ is one-one but not onto.

Question 3

Check the injectivity and surjectivity of the following functions:

i. $\quad f: N \rightarrow N$ given by $f(x)=x^{2}$

ii. $\quad f: Z \rightarrow Z$ given by $f(x)=x^{2}$

iii. $\quad f: R \rightarrow R$ given by $f(x)=x^{2}$

iv. $\quad f: N \rightarrow N$ given by $f(x)=x^{3}$

v. $\quad f: Z \rightarrow Z$ given by $f(x)=x^{3}$

Accepted Answer

i. For $f: N ightarrow N$ given by $f(x)=x^{2} x, y \in N f(x)=f(y) \Rightarrow x^{2}=y^{2} \Rightarrow x=y$

$	herefore f$ is injective.

$2 \in N$. But, there does not exist any $x$ in $N$ such that $f(x)=x^{2}=2$

$	herefore f$ is not surjective

Function $f$ is injective but not surjective.

ii. $\quad f: Z ightarrow Z$ given by $f(x)=x^{2}$

$f(-1)=f(1)=1$ but $-1 
eq 1$

$	herefore f$ is not injective.

$-2 \in Z$ But, there does not exist any $x \in Z$ such that $f(x)=-2 \Rightarrow x^{2}=-2$

$	herefore f$ is not surjective.

Function $f$ is neither injective nor surjective.

iii. $\quad f: R ightarrow R$ given by $f(x)=x^{2}$

$f(-1)=f(1)=1$ but $-1 
eq 1$

$	herefore f$ is not injective.

$-2 \in Z$ But, there does not exist any $x \in Z$ such that $f(x)=-2 \Rightarrow x^{2}=-2$

$	herefore f$ is not surjective.

Function $f$ is neither injective nor surjective.

iv. $\quad f: N ightarrow N$ given by $f(x)=x^{3}$

$x, y \in N$

$f(x)=f(y) \Rightarrow x^{3}=y^{3} \Rightarrow x=y$

$	herefore f$ is injective.

$2 \in N$. But, there does not exist any $x$ in $N$ such that $f(x)=x^{3}=2$

$	herefore f$ is not surjective

Function $f$ is injective but not surjective.

v. $\quad f: Z ightarrow Z$ given by $f(x)=x^{3}$

$x, y \in Z$

$f(x)=f(y) \Rightarrow x^{3}=y^{3} \Rightarrow x=y$

$	herefore f$ is injective.

$2 \in Z$. But, there does not exist any $x$ in $Z$ such that $f(x)=x^{3}=2$

$	herefore f$ is not surjective.

Function $f$ is injective but not surjective.

Question 4

Prove that the greatest integer function $f: R \rightarrow R$ given by $f(x)=[x]$ is neither one-one nor onto, where $[x]$ denotes the greatest integer less than or equal to $x$.

Accepted Answer

$f: R ightarrow R$ given by $f(x)=[x]$

$f(1.2)=[1.2]=1, f(1.9)=[1.9]=1$

$	herefore f(1.2)=f(1.9)$, but $1.2 
eq 1.9$

$	herefore f$ is not one-one.

Consider $0.7 \in R$

$f(x)=[x]$ is an integer. There does not exist any element $x \in R$ such that $f(x)=0.7$

$	herefore f$ is not onto.

The greatest integer function is neither one-one nor onto.

Question 5

Show that the modulus function $f: R \rightarrow R$ given by $f(x)=|x|$ is neither one-one nor onto, where $|x|$ is $x$, if $x$ is positive or 0 and $|x|$ is $-x$, if $x$ is negative.

Accepted Answer

$f: R ightarrow R$ is

$$

f(x)=|x|=\left\{\begin{array}{l}

\mathrm{x}, 	ext { if } \mathrm{x} \geq 0 \

-x, 	ext { if } \mathrm{x}<0

\end{array}ight\}

$$

$f(-1)=|-1|=1$ and $f(1)=|1|=1$

$	herefore f(-1)=f(1)$ but $-1 
eq 1$

$	herefore f$ is not one-one.

Consider $-1 \in R$

$f(x)=|x|$ is non-negative. There exist any element $x$ in domain $R$ such that $f(x)=|x|=-1$

$	herefore f$ is not onto.

The modulus function is neither one-one nor onto.

Question 6

Show that the signum function $f: R ightarrow R$ given by $\quad f(x)=\left\{\begin{array}{l}1, 	ext { if } \mathrm{x}>0 \ 0, 	ext { if } \mathrm{x}=0 \ -1, 	ext { if } \mathrm{x}<0\end{array}ight\}$ is neither one-one nor onto.

Accepted Answer

$f: R ightarrow R$ is $\quad f(x)=\left\{\begin{array}{l}1, 	ext { if } \mathrm{x}>0 \ 0, 	ext { if } \mathrm{x}=0 \ -1, 	ext { if } \mathrm{x}<0\end{array}ight\}$

$f(1)=f(2)=1$, but $1 
eq 2$

$	herefore f$ is not one-one.

$f(x)$ takes only 3 values $(1,0,-1)$ for the element -2 in co-domain

R , there does not exist any $x$ in domain R such that $f(x)=-2$.

$	herefore f$ is not onto.

The signum function is neither one-one nor onto.

Question 7

Let $A=\{1,2,3\}, B=\{4,5,6,7\}$ and let $f=\{(1,4),(2,5),(3,6)\}$ be a function from $A$ to B. Show that f is one-one.

Accepted Answer

$A=\{1,2,3\}, B=\{4,5,6,7\}$

$f: A ightarrow B$ is defined as $f=\{(1,4),(2,5),(3,6)\}$

$	herefore f(1)=4, f(2)=5, f(3)=6$

It is seen that the images of distinct elements of $A$ under $f$ are distinct.

$	herefore f$ is one-one.

Question 8

In each of the following cases, state whether the function is one-one, onto or bijective. Justify your answer.

i. $\quad f: R \rightarrow R$ defined by $f(x)=3-4 x$

ii. $\quad f: R \rightarrow R$ defined by $f(x)=1+x^{2}$

Accepted Answer

i. $\quad f: R ightarrow R$ defined by $f(x)=3-4 x$

$x_{1}, x_{2} \in R$ such that $f\left(x_{1}ight)=f\left(x_{2}ight)$

$\Rightarrow 3-4 x_{1}=3-4 x_{2}$

$\Rightarrow-4 x=-4 x_{2}$

$\Rightarrow x_{1}=x_{2}$

$	herefore f$ is one-one.

For any real number $(y)_{	ext {in }} R$, there exists $\frac{3-y}{4}$ in $R$ such that $f\left(\frac{3-y}{4}ight)=3-4\left(\frac{3-y}{4}ight)=y 	herefore f$ is onto.

Hence, $f$ is bijective.

ii. $\quad f: R ightarrow R$ defined by $f(x)=1+x^{2}$

$x_{1}, x_{2} \in R$ such that $f\left(x_{1}ight)=f\left(x_{2}ight)$

$\Rightarrow 1+x_{1}{ }^{2}=1+x_{2}{ }^{2}$

$\Rightarrow x_{1}{ }^{2}=x_{2}{ }^{2}$

$\Rightarrow x_{1}= \pm x_{2}$

$	herefore f\left(x_{1}ight)=f\left(x_{2}ight)$ does not imply that $x_{1}=x_{2}$

Consider $f(1)=f(-1)=2$

$	herefore f$ is not one-one.

Consider an element -2 in co domain $R$.

It is seen that $f(x)=1+x^{2}$ is positive for all $x \in R$.

$	herefore f$ is not onto.

Hence, $f$ is neither one-one nor onto.

Question 9

Let $A$ and $B$ be sets. Show that $f: A 	imes B ightarrow B 	imes A$ such that $(a, b)=(b, a)$ is a bijective function.

Accepted Answer

$f: A 	imes B ightarrow B 	imes A$ is defined as $(a, b)=(b, a)$.

$\left(a_{1}, b_{1}ight),\left(a_{2}, b_{2}ight) \in A 	imes B$ such that $f\left(a_{1}, b_{1}ight)=f\left(a_{2}, b_{2}ight)$

$\Rightarrow\left(b_{1}, a_{1}ight)=\left(b_{2}, a_{2}ight)$

$\Rightarrow b_{1}=b_{2}$ and $a_{1}=a_{2}$

$\Rightarrow\left(a_{1}, b_{1}ight)=\left(a_{2}, b_{2}ight)$

$	herefore f$ is one-one.

$(b, a) \in B 	imes A$ there exist $(a, b) \in A 	imes B$ such that $f(a, b)=(b, a)$

$	herefore f$ is onto.

$f$ is bijective.

Question 10

$$

f(n)=\left\{\begin{array}{l}

\frac{n+1}{2}, 	ext { if } n 	ext { is odd } \

\frac{n}{2}, 	ext { if } n 	ext { is even }

\end{array}ight\}

$$

Let $f: N ightarrow N$ be defined as for all $n \in N$. State whether the function $f$ is bijective. Justify your answer.

Accepted Answer

$f: N ightarrow N$ be defined as $f(n)=\left\{\begin{array}{l}\frac{n+1}{2}, 	ext { if } n 	ext { is odd } \ \frac{n}{2}, 	ext { if } n 	ext { is even }\end{array}ight\}$ for all $n \in N$.

$f(1)=\frac{1+1}{2}=1$ and $f(2)=\frac{2}{2}=1$

$f(1)=f(2)$, where $1 
eq 2$

$	herefore f$ is not one-one.

Consider a natural number $n$ in co domain $N$.

Case I: $n$ is odd

$	herefore n=2 r+1$ for some $r \in N$ there exists $4 r+1 \in N$ such that

$f(4 r+1)=\frac{4 r+1+1}{2}=2 r+1$

Case II: $n$ is even

$	herefore n=2 r$ for some $r \in N$ there exists $4 r \in N$ such that

$f(4 r)=\frac{4 r}{2}=2 r$

$	herefore f$ is onto.

$f$ is not a bijective function.

Question 11

Let $f: R \rightarrow R$ defined as $f(x)=x^{4}$. Choose the correct answer.

A. $f$ is one-one onto

B. $f$ is many-one onto

C. $f$ is one-one but not onto

D. $f$ is neither one-one nor onto

Accepted Answer

$f: R ightarrow R$ defined as $f(x)=x^{4}$

$x, y \in R_{	ext {such that }} f(x)=f(y)$

$\Rightarrow x^{4}=y^{4}$

$\Rightarrow x= \pm y$

$	herefore f(x)=f(y)$ does not imply that $x=y$.

For example $f(1)=f(-1)=1$

$	herefore f$ is not one-one.

Consider an element 2 in co domain $R$ there does not exist any $x$ in domain $R$ such that $f(x)=2$.

$	herefore f$ is not onto.

Function $f$ is neither one-one nor onto.

The correct answer is D.

Question 12

Let $f: R \rightarrow R$ defined as $f(x)=3 x$. Choose the correct answer.

A. $f$ is one-one onto

B. $f$ is many-one onto

C. $f$ is one-one but not onto

D. $f$ is neither one-one nor onto

Accepted Answer

$f: R ightarrow R$ defined as $f(x)=3 x$

$x, y \in R$ such that $f(x)=f(y)$

$\Rightarrow 3 x=3 y$

$\Rightarrow x=y$

$	herefore f$ is one-one.

For any real number $y$ in co domain R, there exist $\frac{y}{3}$ in R such that $f\left(\frac{y}{3}ight)=3\left(\frac{y}{3}ight)=y 	herefore f$ is onto.

Hence, function $f$ is one-one and onto.

The correct answer is A .

\section*{EXERCISE 1.3}